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Power dissipated in an LCR series circuit connected to an a.c source of emf e is

1. e^2 / √R^2 + (Lo - 1/Co)^2
2. e^2 R / [R^2 + (Lo - 1/Co)^2]
3. e^2 R / √R^2 + (Lo - 1/Co)^2
4. e^2 R / [R^2 + (Lo - 1/Co)^2]

Correct answer: e^2 R / [R^2 + (Lo - 1/Co)^2]

Solution

The power dissipated in an LCR series circuit is given by P = I^2R, where I is the RMS current. Using Ohm's law for AC circuits, I = e / Z, where Z is the impedance. Substituting Z = √(R^2 + (ωL - 1/ωC)^2) into the power formula, we get P = e^2R / [R^2 + (ωL - 1/ωC)^2].

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